Problem: A certain circle can be represented by the following equation. $x^2+y^2-20x+8y+115=0$ What is the center of this circle ? $($
Answer: The strategy We can find the center and radius of a circle by rewriting the given equation in the form of the standard equation of a circle. [What is the standard equation of the circle?] In order to do this, we take the following steps. Complete the square for both the $x^2$ and $y^2$ terms. [How do we complete the square?] Write the equation in the standard form of the circle. Completing the squares $\begin{aligned}x^2+y^2-20x+8y+115&=0\\\\ x^2+y^2-20x+8y&=-115\\\\ (x^2-20x)+(y^2+8y)&=-115 \text{(rearrange terms)}\\\\ (x^2-20x{+100})+(y^2+8y{+16})&=-115{+100}{+16}\end{aligned}$ Notice that we must add ${100}$ and ${16}$ on the right side of the equation, since we added them to the left side of the equation. [How did we get 100 and 16?] Writing the equation in standard form $\begin{aligned}(x^2-20x{+100})+(y^2+8y{+16})&=-115{+100}{+16}\\\\ (x-{10})^2+(y+4)^2&=1\\\\ (x-10)^2+(y-(-4))^2&=1^2\end{aligned}$ Since the equation is now in the standard form, we can conclude that this circle is centered at $(10,-4)$ and has a radius of $1$ unit. Summary The circle is centered at $(10,-4)$. The circle has a radius of $1$ unit.